Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
\[f(xf(y) + x) = xy + f(x)\]for all $x,$ $y.$

Let $n$ be the number of possible values of $f(2),$ and let $s$ be the sum of all possible values of $f(2).$  Find $n \times s.$
Setting $x = 1$ and $y = -1 - f(1),$ we get
\[f(f(-1 - f(1)) + 1) = -1 - f(1) + f(1) = -1.\]Let $a = f(-1 - f(1)) + 1,$ so $f(a) = -1.$

Setting $y = a,$ we get
\[f(0) = ax + f(x).\]Let $b = f(0),$ so $f(x) = -ax + b.$  Substituting into the given functional equation, we get
\[-a(x(-ay + b) + x) + b = xy - ax + b.\]This expands as
\[a^2 xy - (ab + a) x + b = xy - ax + b.\]For this to hold for all $x$ and $y,$ we must have $a^2 = 1,$ and $ab + a = a.$  From $a^2 = 1,$ $a = 1$ or $a = -1.$  For either value, $b = 0.$

Hence, the solutions are $f(x) = x$ and $f(x) = -x.$  Therefore, $n = 2$ and $s = 2 + (-2) = 0,$ so $n \times s = \boxed{0}.$